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Question 1
ప్రశ్న 1
Which formula gives the prismoidal volume between two parallel earthwork cross sections a distance L apart, when their areas are A1 and A2 and the actual mid-sectional area is Am?
Explanation:
The prismoidal rule uses both end-sectional areas and the area at the exact middle of the interval. The end areas are denoted by A1 and A2, while Am denotes the middle sectional area. The full separation between the end sections is L. The required expression is V = L(A1 + 4Am + A2)/6. The middle area is multiplied by four, unlike the two end areas. The formula is exact for a true prismoid and when sectional area varies quadratically with longitudinal distance.
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Question 2
ప్రశ్న 2
The end areas of an earthwork reach are 12 m2 and 24 m2, and its measured mid-sectional area is 18 m2. If the end sections are 30 m apart, what volume is given by the prismoidal rule?
Explanation:
The prismoidal formula is V = L(A1 + 4Am + A2)/6. The end areas are 12 m2 and 24 m2, and the midpoint area is 18 m2. The weighted area sum is 12 + 4 x 18 + 24 = 108 m2. The interval length divided by six is 30/6 = 5 m. Therefore the computed volume is 5 x 108 = 540 m3. Substituting an average of the end areas in place of a separately established midpoint area is unnecessary here.
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Question 3
ప్రశ్న 3
Two survey cross sections used as the ends of a single prismoidal interval are at chainages 0+080 and 0+140. What is the correct chainage for a cross section located exactly halfway between them?
Explanation:
The first chainage represents a distance of 80 m along the alignment. The second chainage represents 140 m from the same origin. Their separation is 140 - 80 = 60 m. Half of this distance is 30 m. Adding 30 m to the first station gives chainage 0+110. The mid-sectional area for the prismoidal rule should refer to this halfway location.
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Question 4
ప్రశ్న 4
For an earthwork interval of 30 m, the end-sectional areas are 10 m2 and 22 m2 and the measured midpoint area is 15 m2. By how much does the TRAPEZOIDAL estimate exceed the PRISMOIDAL estimate?
Explanation:
The trapezoidal estimate uses the mean of the two end-sectional areas. It is 30(10 + 22)/2 = 480 m3. The prismoidal estimate includes the actual midpoint area and is 30(10 + 4 x 15 + 22)/6. This evaluates to 30 x 92/6 = 460 m3. The difference is 480 - 460 = 20 m3. Thus the trapezoidal estimate exceeds the prismoidal estimate by 20 m3 for these supplied sections.
వివరణ:
Question 5
ప్రశ్న 5
An earthwork reach is 30 m long, its end-sectional areas are 10 m2 and 20 m2, and its prismoidal volume is 550 m3. What is the missing mid-sectional area?
Explanation:
The prismoidal formula can be rearranged to determine the missing mid-sectional area. Substituting the given data gives 550 = 30(10 + 4Am + 20)/6. Multiplying both sides by 6/30 gives 110 = 30 + 4Am. Hence 4Am = 80. Dividing by four gives Am = 20 m2. Substitution confirms that 30(10 + 80 + 20)/6 equals 550 m3.
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Question 6
ప్రశ్న 6
The end-sectional areas of an earthwork interval are 8 m2 and 14 m2, and its mid-sectional area is 11 m2. The prismoidal volume is 440 m3. What is the full interval length?
Explanation:
The interval length is obtained by rearranging the prismoidal expression. The weighted area sum is 8 + 4 x 11 + 14 = 66 m2. Hence the volume equation becomes 440 = L x 66/6. Simplifying gives 440 = 11L. The full end-to-end length is therefore L = 40 m. The half-length to the middle section is 20 m, but the formula requires the full 40 m.
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Question 7
ప్రశ్న 7
Two consecutive earthwork intervals are each 20 m long. Their sectional areas in alignment order are 8 m2 at the first end, 12 m2 at the first midpoint, 16 m2 at the common section, 20 m2 at the second midpoint and 24 m2 at the final end. What is the combined volume by applying the prismoidal rule to EACH interval?
Explanation:
The two intervals can be evaluated separately using their respective midpoint areas. For the first interval, the formula gives 20(8 + 4 x 12 + 16)/6 = 240 m3. For the second interval, it gives 20(16 + 4 x 20 + 24)/6 = 400 m3. The area of 16 m2 is used as the end of the first reach and the beginning of the second reach. The combined volume is 240 + 400 = 640 m3. In this particular dataset, applying the prismoidal formula to the entire 40 m using the 16 m2 area at its midpoint also gives 640 m3. Such agreement is not guaranteed for other profiles, so the specified separate-interval procedure should be followed.
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Question 8
ప్రశ్న 8
Five consecutive cross sections at equal 20 m spacings have areas of 6, 10, 14, 18 and 22 m2. Using the COMPOSITE PRISMOIDAL (Simpson's one-third) rule over the full four intervals, what is the total volume?
Explanation:
Four equal intervals allow the direct composite Simpson's one-third expression. With spacing d = 20 m, the area weights are 1, 4, 2, 4 and 1. The weighted sum is 6 + 4 x 10 + 2 x 14 + 4 x 18 + 22 = 168 m2. Multiplying by d/3 gives 20 x 168/3. The computed volume is 1,120 m3. The interior sections are not all assigned the same weight in this rule.
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Question 9
ప్రశ్న 9
Seven consecutive sections have areas of 4, 8, 10, 12, 14, 16 and 20 m2 at equal 15 m spacings. What is their combined volume by the composite prismoidal rule?
Explanation:
Seven sections create six equal intervals, which can be handled using the composite prismoidal rule. The weights on successive sectional areas are 1, 4, 2, 4, 2, 4 and 1. Their weighted sum is 4 + 32 + 20 + 48 + 28 + 64 + 20 = 216 m2. The common spacing is 15 m, so the multiplier is 15/3 = 5 m. The total volume is 5 x 216 = 1,080 m3. A simple unweighted average of all seven sectional areas would not implement the specified rule.
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Question 10
ప్రశ్న 10
One prismoidal reach is 24 m long with end areas 10 m2 and 20 m2 and midpoint area 14 m2. The following reach is 30 m long with end areas 20 m2 and 16 m2 and midpoint area 18 m2. What is the combined volume?
Explanation:
The two reaches have different lengths and must be calculated separately. The first prismoidal volume is 24(10 + 4 x 14 + 20)/6. This equals 24 x 86/6 = 344 m3. The second volume is 30(20 + 4 x 18 + 16)/6 = 540 m3. Their combined volume is 344 + 540 = 884 m3. A uniform-interval composite formula must not be used with these unequal reach lengths.
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Question 11
ప్రశ్న 11
A road embankment has a formation width of 8 m, symmetrical side slopes of 1 horizontal to 1 vertical and level transverse ground. Filling depths vary LINEARLY from 1 m to 3 m across a 30 m reach. What is its prismoidal volume using the end and midpoint cross sections?
Explanation:
For a symmetrical embankment on level ground, the sectional area is A = Bd + md^2. At the first end, A1 = 8 x 1 + 1 x 1^2 = 9 m2. At the midpoint the linear depth is 2 m, giving Am = 8 x 2 + 2^2 = 20 m2. At the final end, A2 = 8 x 3 + 3^2 = 33 m2. The prismoidal volume is 30(9 + 4 x 20 + 33)/6. This gives 30 x 122/6 = 610 m3, which accounts for the quadratic variation of sectional area.
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Question 12
ప్రశ్న 12
An earthwork reach is 24 m long, with end-sectional areas of 9 m2 and 25 m2 and an independently measured midpoint area of 15 m2. By how much does the PRISMOIDAL estimate exceed the MID-SECTIONAL AREA estimate?
Explanation:
The mid-sectional area method multiplies the measured midpoint area by the full reach length. Its estimate is 15 x 24 = 360 m3. The prismoidal method uses both end areas as well as the midpoint area. Its volume is 24(9 + 4 x 15 + 25)/6 = 24 x 94/6 = 376 m3. The difference is 376 - 360 = 16 m3. The methods can disagree because they assign different weightings to the changing cross-sectional areas.
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Question 13
ప్రశ్న 13
A symmetrical embankment has fixed formation width and side slopes of 1.5 horizontal to 1 vertical on level ground. The end depths are 1 m and 3 m, the midpoint depth is their arithmetic mean, and the reach is 24 m long. What is the prismoidal correction, defined as TRAPEZOIDAL volume minus PRISMOIDAL volume?
Explanation:
For a level-ground section with fixed width, symmetrical side slopes and a midpoint depth equal to the mean of the end depths, the prismoidal correction is C = Lm(d2 - d1)^2/6. The side-slope coefficient is m = 1.5 and the length is L = 24 m. The difference in end depths is 3 - 1 = 2 m. Substitution gives C = 24 x 1.5 x 2^2/6 = 24 m3. This correction is positive for the stated geometry and is subtracted from the trapezoidal result. The formula is not a universal correction for arbitrary irregular ground profiles.
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Question 14
ప్రశ్న 14
An embankment on level ground has symmetrical side slopes of 1 horizontal to 1 vertical. Its formation width increases LINEARLY from 6 m to 10 m while the filling depth increases LINEARLY from 1 m to 2 m along a 30 m reach. What is its volume by the prismoidal formula using the midsection obtained from the linearly varying dimensions?
Explanation:
The first sectional area is 6 x 1 + 1^2 = 7 m2. The final sectional area is 10 x 2 + 2^2 = 24 m2. Because width and depth vary linearly, their midpoint values are 8 m and 1.5 m. The midpoint area is 8 x 1.5 + 1.5^2 = 14.25 m2. Applying the prismoidal formula gives V = 30(7 + 4 x 14.25 + 24)/6. The weighted sum is 88 m2, and the volume is 440 m3. Directly averaging the two end areas instead would give the different trapezoidal estimate of 465 m3.
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Question 15
ప్రశ్న 15
A cutting has a constant formation width of 8 m, with side slopes 1 horizontal to 1 vertical on the left and 2 horizontal to 1 vertical on the right. Cutting depth changes LINEARLY from 1 m to 2 m over a 30 m reach on level transverse ground. What is the prismoidal volume?
Explanation:
With unequal side slopes on level transverse ground, sectional area is A = Bd + [(m1 + m2)/2]d^2. The combined side-slope coefficient is (1 + 2)/2 = 1.5. At depths 1 m, 1.5 m and 2 m, the respective areas are 9.5 m2, 15.375 m2 and 22 m2. The midpoint depth of 1.5 m follows from the stated linear change. The prismoidal expression is 30(9.5 + 4 x 15.375 + 22)/6. Its value is 30 x 93/6 = 465 m3.
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Question 16
ప్రశ్న 16
A straight-sided pyramid-like earthwork solid has a base cross-sectional area of 36 m2 at one end and tapers to a point at the other end over a length of 30 m. Its midpoint cross-sectional area is 9 m2, consistent with all transverse linear dimensions halving at the midpoint. What volume is obtained by the prismoidal rule?
Explanation:
The end-sectional areas are 36 m2 and zero because the solid tapers to a point. The mid-sectional area is 9 m2 as expressly stated. The prismoidal formula gives V = 30(36 + 4 x 9 + 0)/6. The weighted sum is 36 + 36 = 72 m2. Therefore the volume is 30 x 72/6 = 360 m3. This also agrees with the pyramid formula of one-third of base area times length, whereas the trapezoidal end-area estimate would be 540 m3.
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Question 17
ప్రశ్న 17
Under which of the following conditions does the prismoidal formula, considered as Simpson's one-third integration rule, give an EXACT volume when the end and midpoint sectional areas are known accurately?
Explanation:
The prismoidal formula is mathematically equivalent to Simpson's one-third numerical integration rule applied to the sectional-area function. Simpson's one-third rule is exact for polynomials of degree three or lower, including constant, linear, quadratic and cubic functions. A true geometrical prismoid normally has a sectional-area function of degree at most two, so the formula is exact for that geometry. The cubic exactness is an additional mathematical property of Simpson's rule. Equal end-sectional areas alone do not guarantee an exact result because the intervening sectional-area variation may be irregular. For higher-degree or arbitrary nonpolynomial functions, the rule generally provides an approximation rather than a guaranteed exact volume.
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Question 18
ప్రశ్న 18
For the DIRECT composite Simpson's one-third form of the prismoidal rule applied to a series of consecutive cross sections, which section arrangement is required without subdividing or treating an interval separately?
Explanation:
The direct composite Simpson's one-third rule operates on pairs of adjacent equal-length intervals. Thus the number of subintervals must be even for an uncomplicated composite application. The corresponding number of cross sections is one greater and is therefore odd. The usual successive weights alternate as 4 and 2 for internal sections, with weight 1 at the ends. Arbitrary unequal spacing does not permit use of that simple composite expression. Separate reach-by-reach prismoidal calculations can still be made when appropriate midpoint areas and lengths are known.
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Question 19
ప్రశ్న 19
Three consecutive sections 15 m apart have areas 10 m2, 16 m2 and 26 m2. The middle section is the exact midpoint of the full 30 m reach. By how much does the COMPOSITE TRAPEZOIDAL volume over the two intervals exceed the PRISMOIDAL volume over the full reach?
Explanation:
The full reach is 30 m because it contains two 15 m subintervals. The composite trapezoidal volume is 15(10 + 16)/2 + 15(16 + 26)/2. These parts are 195 m3 and 315 m3, giving a trapezoidal total of 510 m3. The prismoidal volume over the full reach is 30(10 + 4 x 16 + 26)/6. This equals 30 x 100/6 = 500 m3. Thus the trapezoidal estimate exceeds the prismoidal estimate by 10 m3.
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Question 20
ప్రశ్న 20
At three equally spaced sections 20 m apart, the SOIL excavation areas are 6, 10 and 14 m2, while the ROCK excavation areas are 2, 5 and 8 m2. Using Simpson's one-third form of the prismoidal rule SEPARATELY for each class, what are the soil and rock volumes?
Explanation:
The three cross sections span a total length of 40 m with a midpoint at 20 m. Simpson's one-third form for the two equal intervals is V = (20/3)(A1 + 4Am + A2). For soil, the weighted area sum is 6 + 4 x 10 + 14 = 60 m2. The soil volume is therefore 20 x 60/3 = 400 m3. For rock, the weighted area sum is 2 + 4 x 5 + 8 = 30 m2, giving 20 x 30/3 = 200 m3. Keeping the classes separate preserves the appropriate excavation quantities for costing.
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Question 21
ప్రశ్న 21
A road alignment changes from CUTTING to FILLING at a verified zero-area section at chainage 0+040. The cutting area at 0+000 is 18 m2 and at midpoint 0+020 is 9 m2. The filling area at midpoint 0+055 is 8 m2 and at 0+070 is 12 m2. What separate cutting and filling volumes are obtained by applying the prismoidal rule to the 40 m cutting reach and the 30 m filling reach?
Explanation:
The cutting interval extends from chainage 0+000 to the zero-area station at 0+040. Its midpoint is at chainage 0+020, where the cutting area is 9 m2. The prismoidal cutting volume is 40(18 + 4 x 9 + 0)/6 = 360 m3. The filling interval extends from the zero-area station to chainage 0+070 and has a length of 30 m. Its midpoint is at chainage 0+055, where the filling area is 8 m2. The filling volume is 30(0 + 4 x 8 + 12)/6 = 220 m3. Cutting and filling must be reported separately rather than combined into a net quantity.
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Question 22
ప్రశ్న 22
Over a 30 m earthwork reach, the true measured sectional areas are 10 m2 at the beginning, 15 m2 at the midpoint and 30 m2 at the end. An estimator incorrectly substitutes the ARITHMETIC MEAN of the two end areas for the midpoint area in the prismoidal formula. How does this change the computed volume?
Explanation:
The arithmetic mean of the two end areas is (10 + 30)/2 = 20 m2. The correct prismoidal calculation must instead use the measured midpoint area of 15 m2. With the actual midpoint area, the volume is 30(10 + 4 x 15 + 30)/6 = 500 m3. With the incorrect substituted area of 20 m2, the volume becomes 30(10 + 4 x 20 + 30)/6 = 600 m3. The erroneous result is higher by 600 - 500 = 100 m3. This error effectively reduces the calculation to the trapezoidal estimate and conceals the observed variation in sectional area.
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Question 23
ప్రశ్న 23
For an earthwork reach of 30 m, the end-sectional areas are 12 m2 and 18 m2 and the measured midpoint area is 25 m2. How does the PRISMOIDAL volume compare with the TRAPEZOIDAL volume?
Explanation:
The trapezoidal volume is 30(12 + 18)/2 = 450 m3. The prismoidal rule gives greater weight to the measured midpoint area. Its volume is 30(12 + 4 x 25 + 18)/6. The weighted area sum is 12 + 100 + 18 = 130 m2, giving 650 m3. Therefore the prismoidal volume exceeds the trapezoidal volume by 650 - 450 = 200 m3. This example shows that a prismoidal correction is not always subtractive for an arbitrary measured area profile.
వివరణ:
Question 24
ప్రశ్న 24
Consider three statements. I: The prismoidal formula is exact if sectional area varies quadratically with chainage over the reach. II: The prismoidal and trapezoidal formulas are equal if the measured midpoint area equals the arithmetic mean of the end areas. III: The prismoidal result is always less than the trapezoidal result for any three nonnegative sectional areas. Which option is correct?
Explanation:
Simpson's one-third formula is exact for a quadratic sectional-area function, so Statement I is correct. If Am = (A1 + A2)/2, substituting into the prismoidal expression reduces it algebraically to L(A1 + A2)/2. Therefore Statement II is also correct. The relative sizes of the prismoidal and trapezoidal volumes depend on the actual midpoint area. An unusually large midpoint area can make the prismoidal result greater than the trapezoidal result. Statement III is therefore false, and only Statements I and II are correct.
వివరణ:
Question 25
ప్రశ్న 25
An earthwork computation contains two consecutive PRISMOIDAL reaches with different lengths. For chainages 0+000 to 0+040, the end areas are 8 m2 and 20 m2 and the midpoint area is 14 m2. For chainages 0+040 to 0+070, the end areas are 20 m2 and 10 m2 and the midpoint area is 16 m2. What is the total prismoidal volume?
Explanation:
The first reach length is 40 m, and its weighted sectional-area sum is 8 + 4 x 14 + 20 = 84 m2. Its prismoidal volume is therefore 40 x 84/6 = 560 m3. The second reach is 30 m long, with weighted area sum 20 + 4 x 16 + 10 = 94 m2. Its volume is 30 x 94/6 = 470 m3. Adding the separately calculated reaches gives 560 + 470 = 1,030 m3. The two different reach lengths must not be replaced by a single assumed common interval.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. V = L(A1 + 4Am + A2)/6. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. 540 m3. సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: A. 0+110. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: D. 20 m3. సమాధానం: D.
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Question 5 ప్రశ్న 5Answer: B. 20 m2. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. 40 m. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: C. 640 m3. సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: D. 1,120 m3. సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: A. 1,080 m3. సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: B. 884 m3. సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: C. 610 m3. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: D. 16 m3. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: A. 24 m3. సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: B. 440 m3. సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: C. 465 m3. సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: D. 360 m3. సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: A. The sectional area varies as a polynomial of degree not greater than three with longitudinal distance. సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: B. An even number of equal-length intervals, producing an odd number of cross sections. సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: C. 10 m3. సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: D. 400 m3 of soil and 200 m3 of rock. సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: A. 360 m3 cutting and 220 m3 filling. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: B. It overestimates the correct prismoidal volume by 100 m3. సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: C. The prismoidal volume is greater by 200 m3. సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: A. Statements I and II only. సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: B. 1,030 m3. సమాధానం: B.
